\section{appendix}
\subsection{Proving $T_i$ is a convex func. of $n_i$}\label{Appendix1}
\begin{proof}
A sufficient and necessary condition is that \emph{the second-order derivative $\frac{\partial^2T_i}{\partial n_i^2}$ against $n_i$ is always larger than~0}. Thus, we get $\frac{\partial^2T_i}{\partial n_i^2}$ as follows.
\begin{small}
\begin{equation}\label{SecondOrdDAgainstN}
\begin{aligned}
\frac{\partial^2T_i}{\partial n_i^2}=\frac{fZ_\beta^2t_{\lambda}}{\alpha^2}[e^{\frac{n_i}{f}} (\frac{1}{f^2n_i^2}-\frac{4}{fn_i^3}+\frac{6}{n_i^4})-\frac{6}{n_i^4}]
\end{aligned}
\end{equation}
\end{small}
For simplicity, we denote $(\frac{1}{f^2n_i^2}-\frac{4}{fn_i^3}+\frac{6}{n_i^4})$ as $\Phi$. Then, we have $\frac{\partial^2T_i}{\partial n_i^2}=\frac{fZ_\beta^2t_{\lambda}}{\alpha^2}[e^{\frac{n_i}{f}} \Phi-\frac{6}{n_i^4}]$. We find that $\Phi=(\frac{1}{f^2n_i^2}-\frac{4}{fn_i^3}+\frac{6}{n_i^4})\geq 2\sqrt{\frac{1}{f^2n_i^2}\times\frac{6}{n_i^4}}-\frac{4}{fn_i^3}=\frac{2\sqrt{6}-4}{fn_i^3}>0$. Using the fourth-order Taylor series expansion, it is easy to know $e^{\frac{n_i}{f}}>1+\frac{n_i}{f}+\frac{n_i^2}{2f^2}+\frac{n_i^3}{6f^3}+\frac{n_i^4}{24f^4}$. Since $\Phi>0$, we have:
\begin{small}
\begin{equation}\label{SecondOrdDAgainstNInequality}
\begin{aligned}
\frac{\partial^2T_i}{\partial n_i^2}>\frac{fZ_\beta^2t_{\lambda}}{\alpha^2}[(1+\frac{n_i}{f}+\frac{n_i^2}{2f^2}+\frac{n_i^3}{6f^3}+\frac{n_i^4}{24f^4}) \Phi-\frac{6}{n_i^4}]
\end{aligned}
\end{equation}
\end{small}
Substituting $\Phi=(\frac{1}{f^2n_i^2}-\frac{4}{fn_i^3}+\frac{6}{n_i^4})$ into, and simplifying this inequality, we have $\frac{\partial^2T_i}{\partial n_i^2}>\frac{fZ_\beta^2t_{\lambda}}{\alpha^2}(\frac{2}{fn_i^3}+\frac{1}{12f^4}+\frac{n_i^2}{24f^6})>0$. Since the second-order derivative is always larger than 0, $T_i$ is proved to be a convex function with respect to $n_i$.
\end{proof}


\subsection{Proving $\max\{T_x,T_y\}$ is a convex func. of $f$}\label{Appendix2}
\begin{proof}
We use $\frac{\partial^2T_i}{\partial f^2}$ to denote the second-order derivative of $T_i$ against $f$, which is given as follows.
\begin{small}
\begin{equation}\label{SecondOrdDeriAgainstf}
\begin{aligned}
\frac{\partial^2T_i}{\partial f^2}=\frac{Z_\beta^2t_{\lambda}}{\alpha^2n_i^2}[e^{\frac{n_i}{f}}(\frac{n_i^2}{f^2}-\frac{2n_i}{f}+2)-2]
\end{aligned}
\end{equation}
\end{small}
For clarity, we denote $\frac{n_i^2}{f^2}-\frac{2n_i}{f}+2$ as $\Psi$. Thus, $\frac{\partial^2T_i}{\partial f^2}=\frac{Z_\beta^2}{\alpha^2n_i^2}[e^{\frac{n_i}{f}}\Psi-2]$. We find that $\Psi=\frac{n_i^2}{f^2}-\frac{2n_i}{f}+2=(\frac{n_i}{f}-1)^2+1>0$. Since $e^{\frac{n_i}{f}}>1+\frac{n_i}{f}+\frac{n_i^2}{2f^2}+\frac{n_i^3}{6f^3}+\frac{n_i^4}{24f^4}$, we have:
\begin{small}
\begin{equation}\label{SecondOrdDeriAgainstfInequ}
\begin{aligned}
\frac{\partial^2T_i}{\partial f^2}>\frac{Z_\beta^2t_{\lambda}}{\alpha^2n_i^2}[(1+\frac{n_i}{f}+\frac{n_i^2}{2f^2}+\frac{n_i^3}{6f^3}+\frac{n_i^4}{24f^4})\Psi-2]
\end{aligned}
\end{equation}
\end{small}
Substituting $\Psi=\frac{n_i^2}{f^2}-\frac{2n_i}{f}+2$ into Eq.~(\ref{SecondOrdDeriAgainstfInequ}), we have:
\begin{small}
\begin{equation}\label{InequResult}
\begin{aligned}
\frac{\partial^2T_i}{\partial f^2}>\frac{Z_\beta^2t_{\lambda}}{\alpha^2n_i^2}(\frac{n_i^3}{3f^3}+\frac{n_i^4}{4f^4}+\frac{n_i^5}{12f^5}+\frac{n_i^6}{24f^6})>0
\end{aligned}
\end{equation}
\end{small}
So far, we have proved that $T_i$ is a convex function with respect to $f$, $i\in[1,c]$. Since $T_x$ and $T_y$ are both convex functions with respect to $f$, $\max\{T_x,T_y\}$ is also a convex function against $f$. \end{proof}

